Question: A right cylindrical oil tank is $15$ feet tall and its circular bases have diameters of $4$ feet each. When the tank is lying flat on its side (not on one of the circular ends), the oil inside is $3$ feet deep. How deep, in feet, would the oil have been if the tank had been standing upright on one of its bases? Express your answer as a decimal to the nearest tenth.
Since the oil is $3$ feet deep, we want to find the ratio of the area of the part of the circle covered with oil (part under the horizontal line of the figure below) to the entire area of the circle.
[asy]
draw(Circle((0,0),2));
draw((-1.732,1)--(1.732,1));
draw((0,0)--(-1.732,1));
draw((0,0)--(1.732,1));
draw((0,0)--(0,1));
[/asy] The two radii drawn makes a $120$ degree angle, so the area of the fraction of the circle covered by the oil is $\frac23$ of the circle in addition to the isosceles triangle. We can find the length of half the base of the isosceles triangle by using the Pythagorean theorem on the smaller right triangle. Setting half the length of the base to $x$, we have $x^2+1=4$, so $x=\sqrt{3}$ and the length of the base is $2\sqrt3$. Therefore, we have that the area of the triangle is $\frac12 \cdot 1 \cdot 2\sqrt3=\sqrt3$. So, the area of the part of the circle that's covered in oil is $\frac23 \cdot 4\pi + \sqrt3=\frac83\pi+\sqrt3$.

Thus, we have that the oil takes up $\dfrac{\frac83\pi+\sqrt3}{4\pi} \approx \frac{10.11}{12.57} \approx 0.805$ of the cylinder.

With the cylinder upright, the fraction of the cylinder the oil covers is the same as the fraction of the height the oil covers. Therefore, the oil would be $15 \text{ feet} \cdot 0.805 \approx 12.08 \approx \boxed{12.1}$.